Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int dr[] = {0, 1, 0, -1};
int dc[] = {1, 0, -1, 0};
vector<pair<int, int> > v;
int g[110][110];
int row[110];
int col[110];
int row1[110];
int col1[110];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) scanf("%d", &g[i][j]);
  }
  int sum = 0, sum1 = 0;
  for (int i = 0; i < n; ++i) {
    int mn = 501;
    for (int j = 0; j < m; ++j) {
      mn = min(mn, g[i][j]);
    }
    row[i] = mn;
    sum += row[i];
  }
  for (int j = 0; j < m; ++j) {
    int mn = 501;
    for (int i = 0; i < n; ++i) {
      mn = min(mn, g[i][j] - row[i]);
    }
    col[j] = mn;
    sum += col[j];
  }
  bool ans = true;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      if (col[j] + row[i] != g[i][j]) ans = false;
    }
  }
  for (int j = 0; j < m; ++j) {
    int mn = 501;
    for (int i = 0; i < n; ++i) {
      mn = min(mn, g[i][j]);
    }
    col1[j] = mn;
    sum1 += col1[j];
  }
  for (int i = 0; i < n; ++i) {
    int mn = 501;
    for (int j = 0; j < m; ++j) {
      mn = min(mn, g[i][j] - col1[j]);
    }
    row1[i] = mn;
    sum1 += row1[i];
  }
  bool ans1 = true;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      if (col1[j] + row1[i] != g[i][j]) ans1 = false;
    }
  }
  if (ans1 && sum1 < sum) {
    printf("%d\n", sum1);
    for (int i = 0; i < n; ++i)
      while (row1[i] > 0) {
        printf("row %d\n", i + 1);
        row1[i]--;
      }
    for (int i = 0; i < m; ++i)
      while (col1[i] > 0) {
        printf("col %d\n", i + 1);
        col1[i]--;
      }
  } else if (ans) {
    printf("%d\n", sum);
    for (int i = 0; i < n; ++i)
      while (row[i] > 0) {
        printf("row %d\n", i + 1);
        row[i]--;
      }
    for (int i = 0; i < m; ++i)
      while (col[i] > 0) {
        printf("col %d\n", i + 1);
        col[i]--;
      }
  } else
    puts("-1");
}